cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-6 of 6 results.

A352867 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number that has not appeared that shares a factor with a(n-1), a(n-2), and a(n-1)+a(n-2).

Original entry on oeis.org

1, 2, 6, 4, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36, 38, 40, 42, 44, 46, 48, 50, 52, 54, 56, 58, 60, 62, 64, 66, 68, 70, 72, 74, 76, 78, 80, 82, 84, 86, 88, 90, 92, 94, 96, 98, 100, 102, 104, 106, 108, 110, 112, 105, 7, 21, 35, 49, 63, 77, 91, 119, 126, 133, 140, 114, 116
Offset: 1

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Author

Scott R. Shannon, Apr 06 2022

Keywords

Comments

The sequence shows long runs of even terms differing by 2 which are eventually broken by a number with a product of odd primes less than the last even term. The term after such run-breaking terms is often significantly less than the previous terms, leading to the sequence showing abrupt dips in its values. In the first 200000 terms the longest even-numbered run is 106 terms, and it is likely these runs can grow arbitrarily long. Likewise long runs of odd terms also exist, the longest such run being 133 terms in the same range. However unlike the even-numbered runs which increase by 2 each term the odd-numbered runs increase with differing amounts between each term. Between the large dips in value the majority of terms are concentrated along a line with gradient ~ 1.125. See the linked images.
It takes many terms for the primes to appear, e.g. a(166) = 3, a(239) = 5, a(1841) = 23, a(13325) = 61, a(158205) = 191. They do not appear in their natural order.
Other than the first few terms the only fixed point up to 200000 terms is 63. It is possible more exist although this is unknown. The sequence is almost certainly a permutation of the positive integers.

Examples

			a(4) = 4 as a(2)=2, a(3)=6, a(2)+a(3)=8, and 4 is the smallest unused number that shares a factor with 2, 6, and 8.
a(58) = 105 as a(56)=110, a(57)=112, a(56)+a(57)=222, and 105 = 3*5*7 is the smallest unused number that shares a factor with 110, 112, and 222. This breaks a run of fifty-three consecutive even terms differing by 2.
a(59) = 7 as a(57)=112, a(58)=105, a(57)+a(58)=217, and 7 is the smallest unused number that shares a factor with 112, 105, and 217. This is the second prime to appear after a(2)=2.
		

Crossrefs

Programs

A355647 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number that has not yet appeared that has the same number of divisors as the sum a(n-2) + a(n-1).

Original entry on oeis.org

1, 2, 3, 5, 6, 7, 11, 12, 13, 4, 17, 8, 9, 19, 18, 23, 29, 20, 25, 28, 31, 37, 32, 10, 24, 14, 15, 41, 30, 43, 47, 60, 53, 59, 48, 61, 67, 40, 71, 21, 44, 22, 42, 64, 26, 72, 45, 50, 27, 33, 84, 52, 54, 34, 56, 90, 35, 38, 73, 39, 80, 46, 96, 51, 63, 66, 55, 49, 70, 57, 79, 78, 83, 58, 62, 120
Offset: 1

Views

Author

Scott R. Shannon, Jul 12 2022

Keywords

Comments

In the first 500000 terms the smallest numbers that have not appeared are 15625, 25600, 28561, 36864. It is unknown if these and all other numbers eventually appear. In the same range on eighty-two occasions a(n) equals the sum of the previous two terms, these values begin 3, 5, 17, 64, 90, 73, 120, 144, 192.
See A355648 for the fixed points.

Examples

			a(5) = 6 as a(3) + a(4) = 3 + 5 = 8 which has four divisors, and 6 is the smallest unused number that has four divisors.
		

Crossrefs

Programs

  • Python
    from sympy import divisor_count
    from itertools import count, islice
    def agen():
        anm1, an, mink, seen = 1, 2, 3, {1, 2}
        yield 1
        for n in count(2):
            yield an
            k, target = mink, divisor_count(anm1+an)
            while k in seen or divisor_count(k) != target: k += 1
            while mink in seen: mink += 1
            anm1, an = an, k
            seen.add(an)
    print(list(islice(agen(), 76))) # Michael S. Branicky, Jul 26 2022

A355636 a(1) = a(2) = 1; for n > 2, a(n) is the smallest positive number that has not yet appeared that has the same number of divisors as the sum a(n-2) + a(n-1) but does not equal the sum.

Original entry on oeis.org

1, 1, 3, 9, 18, 6, 30, 100, 24, 12, 196, 48, 20, 28, 80, 60, 72, 84, 90, 40, 42, 8, 32, 54, 10, 729, 2, 14, 81, 15, 108, 21, 22, 5, 26, 7, 27, 33, 96, 34, 56, 126, 66, 320, 35, 38, 11, 4, 39, 13, 44, 46, 132, 51, 55, 57, 162, 58, 140, 150, 70, 156, 62, 65, 17, 69, 74, 77, 19, 160, 23, 82, 78
Offset: 1

Views

Author

Scott R. Shannon, Jul 11 2022

Keywords

Comments

In the first 250000 terms the smallest numbers that have not appeared are 64, 1024, 11664, 15625. It is unknown if these and all other numbers eventually appear.
See A355637 for the fixed points.

Examples

			a(6) = 6 as a(4) + a(5) = 9 + 18 = 27 which has four divisors, and 6 is the smallest unused number that does not equal 27 and has four divisors.
		

Crossrefs

Programs

  • PARI
    listm(nn) = my(va = vector(nn)); va[1] = 1; va[2] = 1; my(m = Map()); mapput(m, 1, 1); for (n=3, nn, my(s=va[n-2]+va[n-1], d=numdiv(s), k=1, vs=Vec(va, n-1)); while (mapisdefined(m, k) || (k==s) || (numdiv(k)!=d), k++); va[n] = k; mapput(m, k, n);); va; \\ Michel Marcus, Jul 11 2022
    
  • Python
    from sympy import divisor_count
    from itertools import count, islice
    def agen():
        anm1, an, mink, seen = 1, 1, 2, {1}
        yield 1
        for n in count(2):
            yield an
            k, target, tsum = mink, divisor_count(anm1+an), anm1+an
            while k in seen or k == tsum or divisor_count(k) != target: k += 1
            while mink in seen: mink += 1
            anm1, an = an, k
            seen.add(an)
    print(list(islice(agen(), 73))) # Michael S. Branicky, Jul 26 2022

A355649 a(1) = a(2) = 1; for n > 2, a(n) is the smallest positive number that has the same number of divisors as the sum a(n-2) + a(n-1).

Original entry on oeis.org

1, 1, 2, 2, 4, 6, 6, 12, 12, 24, 36, 60, 60, 120, 180, 180, 360, 360, 720, 840, 840, 1680, 2520, 2520, 5040, 7560, 10080, 10080, 20160, 27720, 27720, 55440, 83160, 110880, 110880, 221760, 332640, 554400, 554400, 1108800, 1441440, 1441440, 2882880, 4324320, 7207200, 7207200, 14414400, 21621600
Offset: 1

Views

Author

Scott R. Shannon, Jul 12 2022

Keywords

Examples

			a(7) = 6 as a(5) + a(6) = 4 + 6 = 10 which has four divisors, and 6 is the smallest positive number that has four divisors.
		

Crossrefs

A361606 Lexicographically earliest infinite sequence of distinct positive numbers such that, for n > 3, a(n) shares a factor with a(n-1) and a(n-2) but not with a(n-1) + a(n-2).

Original entry on oeis.org

1, 6, 10, 15, 12, 20, 45, 18, 40, 75, 24, 50, 105, 14, 30, 21, 28, 36, 63, 56, 48, 147, 98, 54, 189, 70, 60, 231, 22, 42, 33, 44, 72, 99, 88, 78, 143, 66, 26, 39, 84, 52, 91, 112, 104, 455, 80, 126, 35, 90, 154, 55, 100, 132, 135, 110, 96, 165, 130, 102, 85, 120, 34, 51, 108, 68, 153, 114
Offset: 1

Views

Author

Scott R. Shannon, Mar 17 2023

Keywords

Comments

All terms must contain two or more distinct prime factors. If a(n) was a prime power then a(n+1) would contain the same prime factor, which in turn would imply that a(n) + a(n+1) is a multiple of the prime. But that would make finding a(n+2) impossible as any factor of a(n) would also be a factor of the sum.
To ensure the sequence is infinite a(n) must also contain a prime factor not in a(n-1). If this were not the case the sum a(n-1) + a(n) would be a multiple of the distinct prime factors of a(n), implying a(n+1) would not exist as any factor of a(n) would be a factor of the sum.
The last even term is a(114) = 210. As a(115) = 119 and a(116) = 255, the first occurrence of consecutive odd values, the resulting sum is even, so a(117) must be odd. This forces all subsequent terms to also be odd.
There is a concentration of terms at a(n) ~ 3.4*n. See the linked image. The only fixed point in the first 50000 terms is 14, although it is possible more exist.

Examples

			a(8) = 18 = 2*3*3 as a(6) = 20 = 2*2*5 and a(7) = 45 = 3*3*5 and a(6) + a(7) = 20 + 45 = 65 = 5*13. As the sum contains 5 as a factor a(8) cannot, but it must contain both 2 and 3 while containing a factor not in 45 = 3*3*5. The smallest unused number satisfying these conditions is 18.
		

Crossrefs

Programs

A353075 a(1) = 1, a(2) = 2; for n > 2, a(n) is the smallest positive number that has not appeared that shares a factor with a(n-1) * a(n-2) + |a(n-1) - a(n-2)|.

Original entry on oeis.org

1, 2, 3, 7, 5, 37, 14, 541, 8101, 223, 23, 73, 13, 1009, 11, 12097, 46, 22, 4, 6, 8, 10, 12, 16, 18, 15, 9, 21, 24, 26, 20, 28, 30, 32, 34, 25, 859, 35, 17, 613, 69, 42841, 39, 1713601, 19, 92, 27, 2549, 38, 43, 33, 1429, 115, 44, 42, 36, 40, 48, 50, 52, 54, 45, 51, 57, 60, 49, 65, 55, 63
Offset: 1

Views

Author

Scott R. Shannon, Apr 22 2022

Keywords

Comments

The sequence produces numerous groupings of primes. For example a(3) to a(16) contains thirteen primes in fourteen terms, a(80) to a(102) contains fourteen primes in twenty-three terms. The sequences is conjectured to be a permutation of the positive integers.

Examples

			a(5) = 5 as a(4)*a(3)+|a(4)-a(3)| = 7*3+|7-3| = 25, and 5 is the smallest unused number that shares a factor with 25.
		

Crossrefs

Programs

Showing 1-6 of 6 results.